Answer:
The answers to the questions are as follows
- The temperature of the water bath goes up
- The piston moves out
- Heat flows out of the gaseous mixture
- 173.kJ flows out of the system
Explanation:
- From the question, it is noted that 173.kJ of heat flows out of the system into the insulated water bath therefore the temperature of the water bath goes up
as seen in the relation ΔH = m·c·ΔT
Where ΔH heat measured by temperature rise ΔT of a given mass of water m of specific heat capacity of 4.2 J/g°C
- The amount of heat measured from previous experiment is more than the heat from the present reaction therefore since in the present reaction is constant pressure and from the first law of thermodynamics Energy can neither be created nor destroyed, the balance heat will be transformed to work evidence in the piston moving out
ΣH = Q + W where
w = P × ΔV = P × (P₂ - P₂)
- Heat flows out of the gaseous mixture and is sensed from the rise in the temperature of the water bath
- 173.kJ flows out of the system
Answer:
CHARLES' LAW
given:
= 600 mL = 0.6 L
= 27 °C = 300.15 L
= 77 °C = 350.15 L
conversion:
= 600 mL (1 L / 1000 mL)
= 0.6 L
= 27 °C + 273.15 K
= 300.15 K
= 77 °C + 273.15 K
= 350.15 K
solution:
= ( × ) ÷
= (0.6 L × 350.15 K) ÷ 300.15 K
= 0.7 L
Answer:
Mass of magnesium oxide formed = 35.1 g
Explanation:
Given data:
Mass of Mg = 20.9 g
Mass of O₂ = 15.2 g
Mass of magnesium oxide formed = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg:
Number of moles = mass/molar mass
Number of moles = 20.9 g/ 24 g/mol
Number of moles = 0.87 mol
Number of moles of O₂:
Number of moles = mass/molar mass
Number of moles = 15.2 g/ 32 g/mol
Number of moles = 0.475 mol
Now we will compare the moles of MgO with magnesium and oxygen.
Mg : MgO
2 : 2
0.87 : 0.87
O₂ : MgO
1 : 2
0.475 : 2/1×0.475 = 0.95
Number of moles of MgO formed by Mg are less thus Mg will limiting reactant.
Mass of MgO:
Mass = number of moles × molar mass
Mass = 0.87 mol × 40.3 g/mol
Mass = 35.1 g
<u>Answer:</u>
P2 = 778.05 mm Hg = 1.02 atm
<u>Explanation:</u>
We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.
For this, we would use the equation:
where P1 = 724 mm hg
V1 = 3.05 L
T1 = 298 K
P2 = ?
V2 = 2.6 L
T2 = 173 K
Substituting the given values in the equation to get:
P2 = 778.05 mm Hg = 1.02 atm
